Question

a flagpole is tilted at an angle of 7° from the vertical and toward the sun. it casts a shadow of 14.3 m when the angle of elevation of the sun is 32°. find the length of the flagpole. answer to the nearest tenth of a metre. the answer is: 9.8 m is your answer correct? yes no

Explanation:

Step1: Analyze the triangle angles

The flagpole is tilted \(7^\circ\) from vertical, so the angle between the flagpole and the horizontal (shadow) - related angle: The angle at the flagpole's top with respect to the vertical is \(7^\circ\), so the angle between the flagpole and the horizontal shadow - the angle inside the triangle at the flagpole's base (let's call the flagpole length \(L\), shadow length \(s = 14.3\) m, sun elevation \(32^\circ\)). The angle opposite the shadow: First, the angle between the flagpole and vertical is \(7^\circ\), so the angle between flagpole and horizontal is \(90^\circ - 7^\circ=83^\circ\)? Wait, no. Let's model the triangle: The flagpole (length \(L\)) is tilted \(7^\circ\) toward the sun, so the angle between the flagpole and the vertical is \(7^\circ\), so the angle between the flagpole and the horizontal (shadow) is \(90^\circ - 7^\circ = 83^\circ\)? Wait, no, the sun's elevation is \(32^\circ\) (angle between horizontal and sun's ray). So the triangle has: shadow (side \(s = 14.3\) m), flagpole (side \(L\)), and the sun's ray (side \(R\)). The angles: At the base of the flagpole (on the ground), the angle between shadow and sun's ray is \(32^\circ\) (elevation). The angle at the top of the flagpole: since the flagpole is tilted \(7^\circ\) from vertical toward the sun, the angle between the flagpole and the vertical is \(7^\circ\), so the angle between the flagpole and the sun's ray: Let's calculate the angles in the triangle. The sum of angles in a triangle is \(180^\circ\). The angle at the ground (between shadow and sun's ray) is \(32^\circ\). The angle at the flagpole (between flagpole and shadow): The flagpole is tilted \(7^\circ\) from vertical, so the angle between flagpole and horizontal (shadow) is \(90^\circ - 7^\circ=83^\circ\)? Wait, no, if it's tilted toward the sun, the angle between the flagpole and the vertical is \(7^\circ\), so the angle between the flagpole and the horizontal (shadow) is \(90^\circ - 7^\circ = 83^\circ\), but the sun's elevation is \(32^\circ\), so the angle between the shadow (horizontal) and the sun's ray is \(32^\circ\), so the angle between the flagpole and the sun's ray is \(180^\circ - 83^\circ - 32^\circ=65^\circ\)? Wait, maybe better to use the Law of Sines. Let's define the triangle: Let \(A\) be the top of the flagpole, \(B\) be the base of the flagpole, \(C\) be the end of the shadow. So \(BC = 14.3\) m (shadow), \(AB = L\) (flagpole), \(AC\) is the sun's ray. Angle at \(B\): The flagpole is tilted \(7^\circ\) from vertical, so the angle between \(AB\) and the vertical is \(7^\circ\), so the angle between \(AB\) and horizontal ( \(BC\)) is \(90^\circ - 7^\circ = 83^\circ\). Angle at \(C\): sun's elevation is \(32^\circ\), so angle between \(BC\) (horizontal) and \(AC\) is \(32^\circ\). Therefore, angle at \(A\) is \(180^\circ - 83^\circ - 32^\circ = 65^\circ\). Wait, no, angle at \(B\): if the flagpole is tilted toward the sun, the angle between \(AB\) and the vertical is \(7^\circ\), so the angle between \(AB\) and the sun's ray ( \(AC\)): Wait, maybe I got the angles wrong. Let's re - model: The vertical line is \(V\), flagpole \(AB\) is tilted \(7^\circ\) from \(V\) toward the sun, so \(\angle BAV = 7^\circ\). The sun's elevation is \(32^\circ\), so the angle between horizontal ( \(BC\)) and sun's ray ( \(AC\)) is \(32^\circ\), so \(\angle ACB = 32^\circ\). The angle between vertical \(V\) and horizontal \(BC\) is \(90^\circ\), so the angle between \(AB\) and \(BC\) is \(90^\circ - 7^\circ=83^\circ\) (since \(AB\) is \(7^\circ\) from \(V\) toward \(BC\)). Now, in triangle \(ABC\), angles: \(\angle ABC = 83^\circ\), \(\angle ACB = 32^\circ\), so \(\angle BAC=180 - 83 - 32 = 65^\circ\)? Wait, no, \(\angle ABC\) is the angle at \(B\) between \(AB\) and \(BC\), which is \(90^\circ - 7^\circ = 83^\circ\), \(\angle ACB = 32^\circ\), so \(\angle BAC=180-(83 + 32)=65^\circ\). Now, by Law of Sines: \(\frac{BC}{\sin\angle BAC}=\frac{AB}{\sin\angle ACB}\)

Step2: Apply Law of Sines

We know \(BC = 14.3\) m, \(\angle BAC = 65^\circ\), \(\angle ACB = 32^\circ\), \(AB = L\) (flagpole length). So \(\frac{14.3}{\sin65^\circ}=\frac{L}{\sin32^\circ}\)

First, calculate \(\sin65^\circ\approx0.9063\), \(\sin32^\circ\approx0.5299\)

Then, \(L=\frac{14.3\times\sin32^\circ}{\sin65^\circ}\)

Substitute the values: \(L=\frac{14.3\times0.5299}{0.9063}\)

Calculate numerator: \(14.3\times0.5299\approx14.3\times0.53 = 7.579\) (more accurately, \(14.3\times0.5299 = 14.3\times(0.5 + 0.0299)=7.15+0.42757 = 7.57757\))

Denominator: \(0.9063\)

So \(L=\frac{7.57757}{0.9063}\approx8.36\)? Wait, that's not \(9.8\). Wait, maybe I made a mistake in angle calculation.

Wait, maybe the angle at \(B\) is not \(83^\circ\). Let's re - consider: The flagpole is tilted \(7^\circ\) from vertical toward the sun, so the angle between the flagpole and the vertical is \(7^\circ\), so the angle between the flagpole and the sun's ray: The sun's elevation is \(32^\circ\), so the angle between vertical and sun's ray is \(90^\circ - 32^\circ = 58^\circ\). Since the flagpole is \(7^\circ\) from vertical toward the sun, the angle between flagpole and sun's ray is \(58^\circ - 7^\circ = 51^\circ\)? Wait, no, let's draw the vertical line \(V\), flagpole \(AB\) (tilted \(7^\circ\) from \(V\) towards sun), sun's ray \(AC\) (elevation \(32^\circ\), so angle between \(AC\) and horizontal \(BC\) is \(32^\circ\), so angle between \(AC\) and \(V\) is \(90 - 32 = 58^\circ\). The angle between \(AB\) and \(V\) is \(7^\circ\) (toward \(AC\)), so angle between \(AB\) and \(AC\) is \(58^\circ - 7^\circ = 51^\circ\). Angle at \(C\): \(32^\circ\), angle at \(A\): \(51^\circ\), so angle at \(B\) is \(180 - 51 - 32 = 97^\circ\). Now apply Law of Sines: \(\frac{BC}{\sin\angle BAC}=\frac{AB}{\sin\angle ACB}\)

So \(\frac{14.3}{\sin51^\circ}=\frac{L}{\sin32^\circ}\)

\(\sin51^\circ\approx0.7771\), \(\sin32^\circ\approx0.5299\)

\(L=\frac{14.3\times0.5299}{0.7771}\)

\(14.3\times0.5299\approx7.57757\)

\(L=\frac{7.57757}{0.7771}\approx9.75\approx9.8\) (rounded to nearest tenth)

Ah, so my initial angle calculation was wrong. The correct angle at \(A\) (between flagpole and sun's ray) is \(51^\circ\), angle at \(B\) is \(97^\circ\)? Wait, no, let's re - do the angle analysis:

- Vertical line: \(V\)
- Flagpole: \(AB\), tilted \(7^\circ\) from \(V\) towards the sun, so \(\angle BAV = 7^\circ\)
- Sun's ray: \(AC\), angle of elevation \(32^\circ\), so \(\angle ACB = 32^\circ\) (angle between \(AC\) and horizontal \(BC\))
- Angle between \(V\) and \(BC\) is \(90^\circ\) (since \(V\) is vertical, \(BC\) is horizontal)
- So angle between \(V\) and \(AC\) is \(90^\circ - 32^\circ = 58^\circ\) ( \(\angle VAC = 58^\circ\))
- Then angle between \(AB\) and \(AC\) is \(\angle VAC-\angle BAV=58^\circ - 7^\circ = 51^\circ\) ( \(\angle BAC = 51^\circ\))
- In triangle \(ABC\), sum of angles: \(\angle ABC=180^\circ-\angle BAC - \angle ACB=180 - 51 - 32 = 97^\circ\)
- Now, by Law of Sines: \(\frac{BC}{\sin\angle BAC}=\frac{AB}{\sin\angle ACB}\)
- \(BC = 14.3\) m, \(\angle BAC = 51^\circ\), \(\angle ACB = 32^\circ\)
- \(\sin51^\circ\approx0.7771\), \(\sin32^\circ\approx0.5299\)
- \(AB=\frac{14.3\times\sin32^\circ}{\sin51^\circ}=\frac{14.3\times0.5299}{0.7771}\)
- \(14.3\times0.5299 = 14.3\times0.5+14.3\times0.0299 = 7.15+0.42757 = 7.57757\)
- \(\frac{7.57757}{0.7771}\approx9.75\approx9.8\) (rounded to the nearest tenth)

Answer:

The length of the flagpole is approximately \(\boldsymbol{9.8}\) m, so the answer is correct.